Whovian said:
The analogy can be extended to a 3-dimensional "surface."
Nearly flat doesn't mean "not shaped like a sphere," and there are plenty of other models, such as a 3-torus-like Universe. The 3-sphere is an example.
What happens when you cross that boundary? It should be theoretically possible, and once you do, you're ... umm ... outside the Universe. And the Universe, by definition, is everything, and thus being outside the Universe is impossible. (Again, by definition.)
Whovian and Phinds, great answers! I hope the questioner's curiosity is eventually satisfied.
It's important for people to realize that what is meant by "nearly flat" could in fact be "3-spherical but very large". A very large spherical surface would seem nearly flat to creatures living in it. Near flatness, in that case, just translates into the
3-sphere having a very large radius of curvature.
And a nice thing is that the radius of curvature can be estimated from today's observational data. Alternatively, bounds on the CIRCUMFERENCE of the 3-sphere can be estimated. That is just 2 pi times the radius of curvature. It is how long you would have to travel in a straight line before you get back to your point of departure.
The calculation, in case some readers haven't seen it before, is very simple. You take the Hubble distance (often given as 13.9 billion ly, or 14 billion for round numbers) and multiply by 2 pi. That gives around 88 billion ly. Then you divide that by the square root of minus the Ω
k number that is derived from all the combined data from CMB observations, supernovae, galaxy counts etc.
the estimate of Ω
k keeps getting better as more data is collected. A good recent source is Hinshaw et al, which was just revised as of 30 January.
Just google "Hinshaw nine-year"
These are the final results of the 9 year WMAP project. The report is called "WMAP9" for short.
For instance look at their Table 10, on page 20, where they give 95% confidence intervals which depend on which data sets they combined to get the result.
Combining all the data sets (WMAP+eCMB+BAO+H
0 +SNe) gives the results in the rightmost column.
The confidence interval for - Ω
k is [0.0025, 0.0105]
Let's take the smallest thing to divide 88 billion ly by, that will give the largest circumference. The square root of 0.0025 is 0.05. So divide 88/0.05 = 1760 billion ly.
This works as long as the estimate of Ω
k is negative, so when you take minus you get a positive number, and the square root makes sense.
So according to that confidence interval, from Table 10, the largest the circumference could be is 1760 billion ly.
Now what is the SMALLEST it could be? Well the square root of 0.0105 is about 0.1 and dividing by that is like multiplying by 10, so the smallest the circumference could be is around 880 billion ly.
The main thing is we have no evidence for there being "space outside of space" or for there being any boundary to space. So it could be infinite, or it could be something like a 3-sphere. These are the simplest things to imagine and the model builders are normally aiming for the simplest model with the best fit to the data.
Googling "Hinshaw nine-year" gets you
http://arxiv.org/abs/1212.5226
the dataset abbreviations are explained on page 2.
eCMB means "external CMB dataset" from other than WMAP itself, i.e. South Pole Telescope (SPT) and Atacama Cosmology Telescope (ACT). Of course SNe means supernovae data. BAO (baryon acoustic oscillation) basically means counting galaxies at different distances to detect "ripples" in the average bulk matter of the universe.